Given the determinant: \[ \Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix} \] Which of the following statements are correct? (A) \( \Delta = 2(1 - \cos^2 x) \) (B) \( \Delta = 2(2 - \sin^2 x) \) (C) Minimum value of \( \Delta \) is 2 (D) Maximum value of \( \Delta \) is 4 Choose the correct answer from the options given below:
(B), (C) and (D) only
Evaluating the Determinant and Analyzing Statements
This question requires us to calculate the value of a given 3x3 determinant that contains trigonometric terms (\( \cos x \)) and then verify several statements regarding its value and its minimum and maximum possible values.
Step 1: Evaluate the Given Determinant
The determinant is given by:
\( \Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix} \)
We can evaluate this determinant using the cofactor expansion method. Let's expand along the first row (R1):
\( \Delta = 1 \cdot C_{11} + \cos x \cdot C_{12} + 1 \cdot C_{13} \)
where \( C_{ij} = (-1)^{i+j} M_{ij} \) is the cofactor and \( M_{ij} \) is the minor (the determinant of the submatrix obtained by removing row \( i \) and column \( j \)).
\( M_{11} = \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} = (1)(1) - (\cos x)(-\cos x) = 1 + \cos^2 x \)
\( M_{12} = \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} = (-\cos x)(1) - (\cos x)(-1) = -\cos x + \cos x = 0 \)
\( M_{13} = \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix} = (-\cos x)(-\cos x) - (1)(-1) = \cos^2 x + 1 \)
Now, calculate the cofactors:
\( C_{11} = (-1)^{1+1} M_{11} = 1 \cdot (1 + \cos^2 x) = 1 + \cos^2 x \)
\( C_{12} = (-1)^{1+2} M_{12} = -1 \cdot (0) = 0 \)
\( C_{13} = (-1)^{1+3} M_{13} = 1 \cdot (\cos^2 x + 1) = \cos^2 x + 1 \)
Substitute the cofactors back into the expansion formula for \( \Delta \):
\( \Delta = 1 \cdot (1 + \cos^2 x) + \cos x \cdot (0) + 1 \cdot (\cos^2 x + 1) \)
\( \Delta = 1 + \cos^2 x + 0 + \cos^2 x + 1 \)
\( \Delta = 2 + 2\cos^2 x \)
So, the value of the determinant is \( \Delta = 2 + 2\cos^2 x \).
Step 2: Analyze Statement (A)
Statement (A) claims \( \Delta = 2(1 - \cos^2 x) \).
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we know that \( 1 - \cos^2 x = \sin^2 x \).
So, statement (A) is equivalent to claiming \( \Delta = 2\sin^2 x \).
Our calculated \( \Delta \) is \( 2 + 2\cos^2 x \).
Is \( 2 + 2\cos^2 x \) equal to \( 2\sin^2 x \)? Let's use the identity \( \cos^2 x = 1 - \sin^2 x \) in our expression:
\( \Delta = 2 + 2(1 - \sin^2 x) = 2 + 2 - 2\sin^2 x = 4 - 2\sin^2 x \)
Comparing \( 4 - 2\sin^2 x \) with \( 2\sin^2 x \), we see they are not equal for all values of \( x \). For example, if \( x = 0 \), \( \cos^2 x = 1 \) and \( \sin^2 x = 0 \). Our \( \Delta = 2 + 2(1) = 4 \), while \( 2\sin^2 x = 2(0) = 0 \). Since \( 4 \ne 0 \), statement (A) is incorrect.
Step 3: Analyze Statement (B)
Statement (B) claims \( \Delta = 2(2 - \sin^2 x) \).
Our calculated value is \( \Delta = 2 + 2\cos^2 x \).
Let's simplify the expression in statement (B): \( 2(2 - \sin^2 x) = 4 - 2\sin^2 x \).
Now, let's express our calculated \( \Delta \) in terms of \( \sin^2 x \) using \( \cos^2 x = 1 - \sin^2 x \):
\( \Delta = 2 + 2\cos^2 x = 2 + 2(1 - \sin^2 x) = 2 + 2 - 2\sin^2 x = 4 - 2\sin^2 x \)
Our result \( \Delta = 4 - 2\sin^2 x \) matches the expression in statement (B), which is \( 4 - 2\sin^2 x \). Therefore, statement (B) is correct.
Step 4: Analyze Statement (C) - Minimum Value of \( \Delta \)
Statement (C) claims the minimum value of \( \Delta \) is 2.
We have the expression for \( \Delta \): \( \Delta = 2 + 2\cos^2 x \).
To find the minimum value of \( \Delta \), we need to consider the range of \( \cos^2 x \).
For any real value of \( x \), the value of \( \cos x \) is between -1 and 1, i.e., \( -1 \le \cos x \le 1 \).
When we square \( \cos x \), the possible values of \( \cos^2 x \) are between 0 and 1, i.e., \( 0 \le \cos^2 x \le 1 \).
The minimum value of \( \cos^2 x \) is 0. This occurs when \( \cos x = 0 \) (e.g., when \( x = \frac{\pi}{2} \)).
Substitute the minimum value of \( \cos^2 x \) into the expression for \( \Delta \):
\( \text{Minimum } \Delta = 2 + 2 \times (\text{minimum value of } \cos^2 x) = 2 + 2(0) = 2 \)
The minimum value of \( \Delta \) is 2. Thus, statement (C) is correct.
Step 5: Analyze Statement (D) - Maximum Value of \( \Delta \)
Statement (D) claims the maximum value of \( \Delta \) is 4.
We have the expression for \( \Delta \): \( \Delta = 2 + 2\cos^2 x \).
To find the maximum value of \( \Delta \), we need to consider the maximum value of \( \cos^2 x \).
The maximum value of \( \cos^2 x \) is 1. This occurs when \( \cos x = 1 \) (e.g., when \( x = 0 \)) or \( \cos x = -1 \) (e.g., when \( x = \pi \)).
Substitute the maximum value of \( \cos^2 x \) into the expression for \( \Delta \):
\( \text{Maximum } \Delta = 2 + 2 \times (\text{maximum value of } \cos^2 x) = 2 + 2(1) = 4 \)
The maximum value of \( \Delta \) is 4. Thus, statement (D) is correct.
Summary of Statement Analysis
Let's summarize the correctness of each statement:
- Statement (A): \( \Delta = 2(1 - \cos^2 x) \) - Incorrect.
- Statement (B): \( \Delta = 2(2 - \sin^2 x) \) - Correct.
- Statement (C): Minimum value of \( \Delta \) is 2 - Correct.
- Statement (D): Maximum value of \( \Delta \) is 4 - Correct.
The correct statements are (B), (C), and (D).
Revision Table: Determinant Evaluation and Range Analysis
| Concept | How it was Used | Key Result |
|---|---|---|
| Determinant Expansion (3x3) | Calculated the value of the given determinant using cofactor expansion along the first row. | \( \Delta = 2 + 2\cos^2 x \) |
| Trigonometric Identity \( \sin^2 x + \cos^2 x = 1 \) | Used to rewrite the expression for \( \Delta \) and verify statements (A) and (B). | \( \Delta = 4 - 2\sin^2 x \) (Equivalent form) |
| Range of \( \cos^2 x \) | Determined the minimum and maximum values of \( \Delta \) based on the known range \( 0 \le \cos^2 x \le 1 \). | Min \( \cos^2 x = 0 \), Max \( \cos^2 x = 1 \) |
| Minimum Value of \( \Delta \) | Substituted Min \( \cos^2 x = 0 \) into \( \Delta = 2 + 2\cos^2 x \). | Min \( \Delta = 2 + 2(0) = 2 \) |
| Maximum Value of \( \Delta \) | Substituted Max \( \cos^2 x = 1 \) into \( \Delta = 2 + 2\cos^2 x \). | Max \( \Delta = 2 + 2(1) = 4 \) |
Additional Information: Properties of Determinants and Trigonometric Functions
Determinants are scalar values associated with square matrices and are important for understanding matrix invertibility, solving systems of linear equations, and geometrical interpretations like area or volume transformations.
Evaluating a 3x3 determinant can be done by expanding along any row or column. The choice of row or column can simplify calculations if it contains zeros, though in this case, no zeros are present.
Trigonometric functions like cosine and sine are periodic and have specific ranges. Understanding these ranges is vital for finding the minimum and maximum values of expressions involving them.
- The range of \( \cos x \) is [-1, 1]. This means \( -1 \le \cos x \le 1 \).
- Squaring values in [-1, 1] gives values in [0, 1]. So, the range of \( \cos^2 x \) is [0, 1].
- Similarly, the range of \( \sin x \) is [-1, 1], and the range of \( \sin^2 x \) is [0, 1].
These ranges directly impact the range of functions like \( f(x) = a \pm b \cos^2 x \) or \( g(x) = c \pm d \sin^2 x \) where \( a, b, c, d \) are positive constants. For \( \Delta = 2 + 2\cos^2 x \), since \( 0 \le \cos^2 x \le 1 \), the expression \( 2\cos^2 x \) ranges from \( 2(0)=0 \) to \( 2(1)=2 \). Adding 2 to this range gives \( 2+0=2 \) to \( 2+2=4 \), confirming the range of \( \Delta \) is [2, 4].
Important Questions from Determinants
An even number is the determinant of which of the following matrices?
(A) \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\)
(B) \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\)
(C) \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\)
(D) \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\)
Choose the correct answer from the options given below:
Two pipes A and B together can fill a tank in 40 minutes. Pipe A is twice as fast as pipe B. Pipe A alone can fill the tank in :
There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement.
Let X denote the sum of the numbers on the two cards drawn.
Then P(X > 3) is:
A random variable X has the following probability distribution:
X | -2 | -1 | 0 | 1 | 2 -------------------------------------------- P(X) | 0.2 | 0.1 | 0.3 | 0.2 | 0.2
The variance of X will be:
The angle between two lines whose direction ratios are proportional to \( (\sqrt{3} - 1) \), \( (-\sqrt{3} - 1) \), and -4 is:
